Question

A 50.0 kg gorilla is sitting on the limb of a tree 3.2 m above the ground. The gorilla jumps
down from the tree limb to the ground. Use the conservation of energy to find the
velocity of the gorilla just before hitting the ground

Answer 1

Answer:

mass of gorilla = 50kg

height above the ground = 3.2 m

Potential energy possesed by the gorilla at the time of sitting at limb of tree = mgh

where m= mass of gorilla

g= gravitational acceleration pull by earth

h = height above ground

This potential energy will be converted to kinetic energy just before touching the ground

PE= KE

mgh= 1/2 mv2

v= √ 2gh = √ 2 x 10x3.2 = 8m/s

Answer 2

Explanation:

Potential energy of the gorilla when on top of the tree = mgh = 50 x 9.8 x 3.2

By conservation of energy,

Potential energy at top = Kinetic energy at bottom

$mgh \: = \: \frac{1}{2} m {v}^{2}$

the mass gets cancelled out and we get

${v }^{2} \: = \: 2gh$

$v \: = \sqrt{2 \times 9.8 \times 3.2}$

$v = 7.92m {s}^{ - 1}$

hope it helps